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Olin [163]
3 years ago
6

the work of a student to solve a set of equations is shown: equation a: y = 15 − 2z equation b: 2y = 3 − 4z step 1: −2(y) = −2(1

5 − 2z) [equation a is multiplied by −2.] 2y = 3 − 4z [equation b] step 2: −2y = 15 − 2z [equation a in step 1 is simplified.] 2y = 3 − 4z [equation b] step 3: 0 = 18 − 6z [equations in step 2 are added.] step 4: 6z = 18 step 5: z = 3 in which step did the student first make an error? step1 step 2 step 3 step 4
Mathematics
1 answer:
nikdorinn [45]3 years ago
6 0
If you would like to solve a set of equations, you can do this using the following steps:

equation a: y = 15 - 2z       /*(-2)
equation b: 2y = 3 - 4z
__________________
-2y = (-2) * (15 - 2z)
-2y = -30 + 4z
2y = 3 - 4z

The mistake was made at step 2: −2y = 15 − 2z [equation a in step 1 is simplified.]. The correct simplification would be <span>-2y = -30 + 4z. </span>
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